Problem: Find $\lim_{x\to 3}\dfrac{-4}{2x-5}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4$ (Choice B) B $\dfrac{4}{11}$ (Choice C) C $1$ (Choice D) D The limit doesn't exist
Solution: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to 3}\dfrac{-4}{2x-5}&=\dfrac{-4}{2(3)-5} \\\\ &=\dfrac{-4}{6-5} \\\\ &=-4 \end{aligned}$ We got a finite number. Since $\dfrac{-4}{2x-5}$ is continuous across its domain, we can determine that $\lim_{x\to 3}\dfrac{-4}{2x-5}$ is indeed equal to $-4$. In conclusion, $\lim_{x\to 3}\dfrac{-4}{2x-5}=-4$.